Question

the perimeter of a rhombus is 52 cm and one of its diagonals is 10 cm . find the lenght of other diagonal and the area of rhombus

Answer 1

JioGiven :-
perimeter of rhombus =52cm
diagonal =10cm

(as one of the property = all sides of rhombus are equal)
each side = 52/4
each side =13cm
we can name the rhombus ABCD
AB=BC=CD=DA
AC= diagonal 1 = 10cm
draw diagonal 2= BC
(another property of rhombus = diagonals of rhombus bisect each other at 90 degree )
so,
AO = 5 cm
OC = 5cm
we can take AOD triangle
= (AO)^2+(OD)^2 =(AD)^2
= 5^2+(OD)^2 = (13)^2
= 25+(OD)^2 = 169
= (OD)^2 = 169-25
=(OD)^2 = 144
= OD = √144
= OD = 12 cm

now,
BO = OD
BO= 12cm
OD= 12cm
total BD = 12+12=24cm
now we get diagonal 2=24 cm

area of rhombus =
$= \frac{1}{2} \times diagonal1 \times diagonal2 \\ = \frac{1}{2} \times 10 \times 24 \\ = 5 \times 24 \\ = 120 {cm}^{2}$

Answer 2

Perimeter =4xa
52=4xa
52/4=a
13=a
a=13

Let ABCD is a rhombus with diagonals AC and BD which intersect each other at O
AC=10=>AO=5
Let BO =x and AB =13 cm

By Pythagoras theorem

C square= A square + B square
13 square= 5 square+ B square
169 = 25 + B square
B square=169 - 25
B square = 144
B = 12

Therefore BO = 12 cm

Diagonal BD = 2 x 12 = 24 cm

Area = 1/2 ( product of diagonals )
Area = 1 x 10 x 24
Area = 1 x 5 x 24
Area = 120

The length of other diagonal is 24
And area of the rhombus is 120